anonymous
  • anonymous
f(x)=2sin(1/2x)-3 what is the amplitude and the period?
Mathematics
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chestercat
  • chestercat
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anonymous
  • anonymous
amplitude=2. period=4 pi
anonymous
  • anonymous
The amplitude of a sine wave is the absolute value of half difference of the maximum and minimum ordinates of the wave. This function as written does not satisfy those conditions. Hence unless you specify an interval, you cannot specify an amplitude. Moreover, a function is periodic if and only if f(x+a)=f(x). There is no such number for this function. Hence, this function is not periodic either. If this explanation does not make any sense, graph this function paying particular attention where x =0.
anonymous
  • anonymous
P.S. In the interval [-1,1], this function is discontinous and therefore does not have an amplitude there. Outside the interval [-1,1], the function does not repeat itself, hence it has no period there either. I hope I am not confusing you.

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anonymous
  • anonymous
I think he means .5 *x not x/2
anonymous
  • anonymous
lol, in a much simpler way: you know the generasl equation of this right? which is: \[ = A(\sin \omega t + \theta) \] where : A = amplitude w = 2pi/F or in other words w = 2piT, where F is the frequency and T is the period. Since you want to find the period I advice you to use the other equation :) so in this case: f(x) = A(sin wx + theta) = 2(sin (1/2) x ) - 3 - Amplitude = 2 - w = 1/2 plug in w in the equation to find T (period) : \[\omega = 2\pi T\] \[T = \omega/2\pi\] \[T = (1/2)/(2\pi)\] \[T = 1/4\pi \approx 0.08 secs\] Correct me if I'm wrong ^_^
anonymous
  • anonymous
I think you are mostly right, but i think that \[\omega = 2 \Pi F\] instead of \[\omega = 2 \Pi T\] This would make more sense, because having a t of 0.08 secs is a really fast time for period. Also the units match, because omega is in (rad/2) and that means you need to have time on the bottom.
anonymous
  • anonymous
lol, I think so too. It's been a while since I have last studied physics :) thank you though ^_^

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